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What are the necessary conditions of a function for Taylor theorem remainder ?


Definition: Let $f$ be a function with derivatives of orders $1, 2, \dotsc, n$ existing at a point $a$ in the domain. The Taylor polynomial of $f$ at $a$ of order $n$ is: $$T_{n,f,a}(x)=\sum_{k=1}^{n}\dfrac{f^{(n)}(a)}{k!}(x-a)^k.$$


Theorem: Let $f$ be $(n + 1)$ times differentiable on the open interval $(a,b)$ with $f^{(n)}$ continuous on the closed interval $[a,b]$. Then there exits $c \in (a,b)$: $$f(b)=\sum_{k=1}^n\dfrac{f^{(n)}(a)}{k!}(b-a)^k+\dfrac{f^{(n+1)}(c)}{(n+1)!}(b-a)^{(n+1)}.$$


Should these conditions ($(n + 1)$ times differentiable, continuous on the closed interval) always be?

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Yes. Taylor-Lagrange formula comes from the generalisation of the MVT $(n=0) $.

To use MVT, $f $ must be continuous at the colsed $[a,b] $ and differentiable at the open $(a,b) $.

For Taylor-Young formula, you need only that $f^{(n)}(a) $ exists.

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  • $\begingroup$ MVT = Mean value theorem ? $\endgroup$ – Almot1960 Aug 21 '17 at 21:50
  • $\begingroup$ @Almot1960 Yes. the MVT is a consequence of Rolle's Theorem. $\endgroup$ – hamam_Abdallah Aug 21 '17 at 21:50

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